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On 3/25/2025 9:59 AM, dbush wrote:Sure it can, via a representation.On 3/25/2025 9:14 AM, olcott wrote:Because no TM can ever take another actual TM as an input.On 3/25/2025 3:32 AM, joes wrote:>Am Mon, 24 Mar 2025 22:29:06 -0500 schrieb olcott:>On 3/24/2025 10:12 PM, dbush wrote:>On 3/24/2025 10:07 PM, olcott wrote:On 3/24/2025 8:46 PM, André G. Isaak wrote:And not all mathematical functions are computable, such as the haltingOn 2025-03-24 19:33, olcott wrote:https://en.wikipedia.org/wiki/Pure_function Is another way to look atOn 3/24/2025 7:00 PM, André G. Isaak wrote:>Those are called computations or algorithms, not computableIn the post you were responding to I pointed out that computableComputable functions implemented using models of computation would
functions are mathematical objects.
seem to be more concrete than pure math functions.
functions.
computable functions implemented by some concrete model of
computation.
function.You just aren't paying enough attention. Turing machines are never inFalse. *All* turing machine are the domain of the halting function,The halting problems asks whether there *is* an algorithm which canNone-the-less it only has specific elements of its domain as its
compute the halting function, but the halting function itself is a
purely mathematical object which exists prior to, and independent of,
any such algorithm (if one existed).
entire basis. For Turing machines this always means a finite string
that (for example) encodes a specific sequence of moves.
and the existence of UTMs show that all turnBing machines can be
described by a finite string.
the domain of any computable function. <snip>Fine, their descriptions are, and their behaviour is computable ->
by running them.
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Halt deciders
Don't exist, because no H satisfies this requirement:
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Given any algorithm (i.e. a fixed immutable sequence of instructions) X described as <X> with input Y:
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A solution to the halting problem is an algorithm H that computes the following mapping:
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(<X>,Y) maps to 1 if and only if X(Y) halts when executed directly
(<X>,Y) maps to 0 if and only if X(Y) does not halt when executed directly
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